Arithmetic progression

Arithmetic and Geometric progressions.

Arithmetic progression

Postby Math Tutor » Mon Jan 07, 2008 7:13 am

Find the sum of the numbers
1 + 2 + 3 + ... + 100 = ?





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Postby charlie_eppes » Fri Dec 19, 2008 10:09 am

Find [tex]a_{1}[/tex] and "d" in this arithmetic progression.
[tex]|a_{5}:a_{3}=4[/tex]
[tex]|a_{2}.a_{6}=-11[/tex]
What about this?

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Postby Math Tutor » Fri Dec 19, 2008 10:25 am

What is the value of [tex]a_5[/tex] and [tex]a_2[/tex],
[tex]a_2[/tex] and [tex]a_6[/tex],
using [tex]a_1[/tex] and [tex]d[/tex]?

http://www.math10.com/en/algebra/arithm ... ssion.html

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Postby charlie_eppes » Fri Dec 19, 2008 11:56 am

[tex]|{a_{5}}^{2}-{a_{3}}^{2}=72[/tex]
[tex]|a_{5}+a_{8}=28[/tex]
Something is wrong in my answers. I can't define it.
And this
[tex]|5a_{1}+10a_{5}=0[/tex]
[tex]|S_{4}=14[/tex]

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Postby charlie_eppes » Fri Dec 19, 2008 12:10 pm

[quote="charlie_eppes"][tex]|{a_{5}}^{2}-{a_{3}}^{2}=72[/tex]
[tex]|a_{5}+a_{8}=28[/tex]

[tex]|(a_{1}+4d)^{2}-(a_{1}+2d)^{2}=72[/tex]
[tex]|a_{1}+4d+a_{1}+7d=28[/tex]

[tex]2a_{1}+11d=28[/tex]
[tex]a_{1}[/tex] ?

[tex]{a_{1}}^{2}+8a_{1}d+16d^{2}-{a_{1}}^{2}+4a_{1}d-4d^{2}=72[/tex]
[tex]12d^{2}+12a_{1}d=72[/tex]
[tex]d^{2}+a_{1}d=6[/tex]
Then?

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Postby Math Tutor » Fri Dec 19, 2008 5:36 pm

From the first equation :
[tex]a_1 = \frac{(28-11d)}{2}[/tex]
[tex]a_1 = 14-5.5d[/tex]
then solve the quadratic equation and find d

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Postby charlie_eppes » Fri Dec 19, 2008 5:48 pm

teacher wrote:From the first equation :
[tex]a_1 = \frac{(28-11d)}{2}[/tex]
[tex]a_1 = 14-5.5d[/tex]
then solve the quadratic equation and find d

Why [tex]a_1 = 14-5.5d[/tex]
How to solve [tex]d^2 +a_1 d=6[/tex]

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Postby charlie_eppes » Sat Dec 20, 2008 2:15 pm

charlie_eppes wrote:
teacher wrote:From the first equation :
[tex]a_1 = \frac{(28-11d)}{2}[/tex]
[tex]a_1 = 14-5.5d[/tex]
then solve the quadratic equation and find d

10x!

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sending solution

Postby parvez » Mon Jun 01, 2009 11:43 am

:lol: for this arthimetic progression,
sum=n(n+1)/2
sum=100(100+1)/2
sum=5050 :lol:

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Postby shrinidhi » Wed Aug 19, 2009 6:55 am

a=1
d=1
n=100

s=100/2[2*1+(100-1)*1]
=50[2+99]
=50[101]
= 5050[/u]

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Postby MV RAO » Thu Dec 10, 2009 1:57 am

charlie_eppes wrote:Find [tex]a_{1}[/tex] and "d" in this arithmetic progression.
[tex]|a_{5}:a_{3}=4[/tex]
[tex]|a_{2}.a_{6}=-11[/tex]
What about this?


[tex]a_{2} = a_{1} + d; a_{3} = a_{1} +2d, a_{5} = a_{1} + 4d, a6 = a1 + 5d[/tex]

[tex]a_{5}/a_{3} = (a_{1} + 4d)/(a_{1} + 2d) = 4 or 4a_{1} + 8d = a_{1} + 4d or 3a_{1} = -4d or a_{1} = -4d/3[/tex]
[tex](a_{1} + d) (a_{1} + 5d) = (-4d/3 + d)(-4d/3 + 5d) = (-d/3_)11d/3) = -(11d^2)/9 = -11 or d^2 = 8[/tex]
So d=3 and [tex]a_{1} = -4[/tex]

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Postby MV RAO » Thu Dec 10, 2009 2:31 am

charlie_eppes wrote:[tex]|{a_{5}}^{2}-{a_{3}}^{2}=72[/tex]
[tex]|a_{5}+a_{8}=28[/tex]
Something is wrong in my answers. I can't define it.
And this
[tex]|5a_{1}+10a_{5}=0[/tex]
[tex]|S_{4}=14[/tex]


a5^2 – a3^2 = 72, a5 + a8 = 28, find a1 and d
(a1 + 4d) + (a1 + 7d) = 2a1 + 11d = 28 or a1 = (28 – 11d)/2
(a1 + 4d)^2 – (a1 + 2d)^2 = a1^2 + 16d^2 + 8a1d – a1^2 – 4d^2 – 4a1d = 12d^2 + 4a1d = a2d^2 + 4d(28 – 11d)/2 = 12d^2 + 56d – 22d^2 = 72
Or -10d^2 + 56d = 72 or 5d^2 – 28d + 36 = (5d – 18 )(d – 2) = 0 or d = 2
a1 = (28 – 11d)/2 = 14 – 11 = 3

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Postby joetraff » Fri Aug 13, 2010 12:08 pm

Hey Auster !!
Is it some kinda mathematical prank ? :P

Thanks
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Re: Arithmetic progression

Postby cik are u » Fri Feb 25, 2011 12:05 am

the sum of the first n term of arithmetic progression is given by Sn = n^2+3n
a) the first term
b) the second term
c)the common difference
how to solve this question??????

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Re: Arithmetic progression

Postby Math Tutor » Fri Feb 25, 2011 4:16 am

The problem seems to be incomplete.
Check up once again if the condition is correct, please.

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Re: Arithmetic progression

Postby sajid121 » Thu Apr 26, 2012 6:08 am

Need help to solve this problem

Postby Auster1989 » Mon Aug 09, 2010 10:53 pm
hello,

My name is Auster from canada,Can you help me guys to solve this problem?

Think of a number between eight-hundred and ten and nine-hundred and sixty-three, then add one thousand and fifty-three.

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Re: Arithmetic progression

Postby sudarshan » Wed Jul 18, 2012 9:57 pm

Find the sum of the numbers
1 + 2 + 3 + ... + 100 = ?


1st method
First term(a)=1
Last term(b)=100
Numbers of terms(n)=100
Using formula
Sum=n/2(a+b)
=100/2(1+100)
=50*101
=5050

Alternative method
First term(a)=1
Common difference(d)=1
Numbers of terms(n)=100
Using formula
sum=n/2(2a+(n-1)d)
=100/2(2*1+(100-1)*1)
=50(2+99)
=50*101
=5050
Hence, ans=5050

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Re: Arithmetic progression

Postby Anne20 » Fri Aug 24, 2012 1:50 pm

n(n+1)/2 = 100(100 + 1)/ 2 = 100 x 101 / 2 = 5050

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Re: Arithmetic progression

Postby leesajohnson » Fri Jan 29, 2016 6:15 am

Let n= 100
so puttung value of n into n(n+1)/2

=> 100(100+1)/2

=> 100(101)/2

=> 10100/2

=> 5050

The sum of the given numbers is 5050.

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Re: Arithmetic progression

Postby Guest » Sat Apr 30, 2016 12:34 pm

Given Arithmetic progression series had common difference as 1. hence it's sum till 100th term is
n(n+1)/2
= 100(100 + 1)/ 2
= 100 x 101 / 2
= 5050
http://www.techcrashcourse.com/2015/08/c-program-generate-arithmetic-series-sum.html

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